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I'm working on a game theory problem I can't seem to figure out.

Players 1 and 2 are bargaining over how to split $\$10$. Each player names an amount $s_i$, between 0 and 10 for herself. These numbers do not have to be in whole dollar units. The choices are made simultaneously. Each player's payoff is equal to her own money payoff. In all the cases below, if $s_1+s_2\leq 10$, then the players get the amounts they named (and the remainder is destroyed).

(a) In the first case, if $s_1+s_2 >10$, then both players get zero and the money is destroyed.

(b) In the second case, if $s_1+s_2 >10$ and the amounts named are different, then the person who names the smaller amount gets that amount and the other person gets the remaining money. If $s_1+s_2 >10$ and $s_1=s_2$, then both players get $\$5$.

(c) In the third case, the games (a) and (b) must be played such that only integer dollar amounts can be named by both players.

Determine the pure strategy Nash Equilibria for all games (a) - (c).

I'm pretty sure I've figured out (a), noting that if player 2 chooses a strategy such that $0 \leq s_2 < 10$, the best response of player 1 is $BR_1(s_2) = 10- s_2$. If $s_2 = 10$, then $BR_1(s_2) = [0,10]$ since the payoff is $0$ regardless. The same holds for player 2. The Nash Equilibria is the intersection of the BR lines, and occur at $(10,10)$ and the line $s_1+s_2=10$.

For (b), my thought is that if player 2 chooses $s_2 \leq 5$, then $BR_1(s_2) = 10-s_2$ as in (a). However, if $s_2 > 5$, I feel that $BR_1(s_2) = s_2 -\epsilon$ for some very small $\epsilon >0$. This way, the total amount will be over $\$10$, but player 1 will have the smaller amount and thus get his money. However, I'm not sure if this is right or how to find the Nash Equilibrium in this case. Any help would be greatly appreciated.

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This was just bumped to the homepage by Community ♦ because it had only unvoted answers. Since it turns out that these were wrong, here's a new one:

In (b), every strategy below $\$5$ is strictly dominated by $\$5$, since that always wins exactly $\$5$ independent of the other player's strategy.

On the other hand, no strategy above $\$5$ has a best response. If $s_1\gt5$, then if $s_2\ge s_1$, Player $2$ can improve by switching to $s_2=s_1-\epsilon$; whereas if $s_2\lt s_1$, Player $2$ can improve by switching to $\frac{s_1+s_2}2$.

Thus the only pure-strategy Nash equilibrium is $(5,5)$, where deviating in either direction reduces the payoff.

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So I'm working through Ben Polack's "Game Theory" course (free from Open Yale Courses, by the way, and highly recommended). He has an identical problem in problem set 2, number 4.

(Disclaimer: I only have a BS in pure math, so I could be missing something here but I'll take a crack at this.)

As I understand Nash Equilibrium (NE), it's when each player's response is a best response to every other player's response. More formally, for a strategy profile $(s_1^*, s_2^*,...,s_N^*)$, it is a NE if and only if for each player $i \in \{ 1, 2 , ... , N \}$ her choice $s_i^*$ is a best response to $s_{-i}^*$. (Sidenote about notation: Polack uses $s_{-i}^*$ to denotes all strategies besides the strategy of player $i$. I'm not sure how common this notation is, so I thought I'd point it out.)

Since we're only dealing with two players, we only need player 1's strategies that are best responses to player 2's best responses.

Unfortunately, I have a slight nit-pick on the OP's 4(a) answer, so I'll address that first.

4(a): It is true that the line $s_1 + s_2 = 10$ can give NE's. But I think it's more precise to say that they're strategy profiles of the form $(s_1, 10-s_1)$ or equivalently $(10-s_2, s_2)$ for $s_1,s_2 \in [0,10)$. It's also true that if $s_2 = 10$ then $BR_1(s_2)=x$ where $x \in [0,10]$. By symmetry, if $s_1 = 10$ the $BR_2(s_1)=y$ where $y \in [0,10]$. So, like the OP said, there's another NE for the strategy profile $(10, 10)$.

4(b): joriki makes some really good points on this in the continuous case, so read his answer, too. I'll address this in the discrete case (specifically if we're dealing with cents). It's not specified in the problem set if fractional cents are allowed, but I'll go with the assumption that there are not fractional cents.

There are two cases to think about:

(1) $s_1 + s_2 \leq 10$

(2) $s_1 + s_2 > 10$

Like joriki argues, any choices by player 1 where $s_1 < 5$ are strictly dominated by $s_1'=5$. Why? Because in case (1), the payoff of player 1 is $u_1(5,s_2) = \$5$ since both players receive a payoff of whatever number they choose. In case (2), $s_2 > 5$ so player one gets a payoff of the lower number. Thus, it's still true that $u_1(5,s_2)= \$5$.

But what about $s_1 > 5$ in the discrete case using cents? Well, this is were things become a bit different. You end up with NE's for the strategy profiles $(5,5)$, $(5,5.01)$, and $(5.01,5)$, and $(5.01,5.01)$.

It might be easier see this if we start with a case where there isn't a best response from both players, so let's consider when player 2 chooses $s_2=5.02$. So what's player 1's best response? To slightly undercut player 2's choice by a penny (that devious devil). So $BR_1(5.02)=5.01$ which yields $u_1(5.01,5.02)= \$5.01$ and $u_2(5.01,5.02)=\$4.99$.

So what would be player 2's best response to $s_1=5.01$? A couple options, actually.

Player 2 could pick $s_2=5.01$ and they'd both get a payoff of $u_1=\$5$ and $u_2=\$5$ since they both get a payoff of \$5 when $s_1+s_2>10$ and $s_1=s_2$.

Another choice player 2 could make is $s_2=5.00$ (which would still maximize $u_2$ against $s_1=5.01$). Why? Because $u_2=\$5$ since it's the smaller number and the player who chooses the smaller value gets that amount of money as their payoff.

Hence, player 2 has two best responses for $s_1=5.01$. The argument is the same for player one, so we have NE's $(5,5.01)$, $(5.01,5)$, and $(5.01,5.01)$.

Lastly, similar to joriki's argument, if $s_1=5$ then the way for player 2 to maximize their utility function (that is, their best response) is to pick 5 or higher. The argument is symmetric for player 1, and the only overlapping best responses are when $s_1=5$ and $s_2=5$, therefore the strategy profile $(5,5)$ is also a NE.

Hope that's helpful!

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I believe the pure equilibrium for (a) is that both parties choose 5 dollars. Knowing that your adversary has chosen 5 dollars, you can either choose more than 5 dollars, in which case you both get nothing, or you choose an amount less than 5 dollars in which case you get less than 5 dollars, or you can choose 5 dollars in which case you both get 5 dollars. Since the game is symmetric, this is the only case where each player does the same thing knowing what the other player's strategy is, and the payoff for the game for each player is the optimal 5 dollars due to symmetry.

The same should be true when you restrict to integer dollar amounts. However I'm not sure about case (b) and also what happens when you restrict that case to integer amounts.

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    $\begingroup$ I agree with OP that answer to a) is any combination $s_1+s_2=10$. If you know he is choosing 9, you should choose 1, and so forth. For b), follow the logic of Nash--if $s_1>5$, what is the best reply, and is $s_1$ still a best reply to that? $\endgroup$ – Trurl Feb 16 '15 at 15:21
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Here, in case b), given the sum >10, a player would always try to choose a pay off that would maximize his utility. He would choose 10, and hence the rest of the amount that is 0 should go to the player 2, provided he had chosen an amount less than 10. However, player 2 also would try to maximize his utility and choose max amount 10. This would lead to an equal pay off of (5,5). This is PSNE.

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  • $\begingroup$ the answer above is incorrect for b). If my opponent picks $10, I will pick $9.99 and he will only get a penny. The NE for b) are (5, 5) (5.01, 5) and (5, 5.01). The are the only situations where neither player can profitably deviate. $\endgroup$ – David Waldstein Oct 10 '17 at 16:54
  • $\begingroup$ @DavidWaldstein (5.01, 5) is not a NE for b) since player 1 is getting 0 and can improve by picking 5 instead. (5, 5) is the only equilibrium for b). $\endgroup$ – Rchn Jul 12 '18 at 9:45

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